Logic behind php if code - php

Question

I'm trying to understand this code in the company's system that I am working in but this doesn't make any sense for me.
$k = 48;
if (($k & 2) > 0) {
echo "2 is true";
}
echo "<br />";
if (($k & 4) > 0) {
echo "4 is true";
}
echo "<br />";
if (($k & 8) > 0) {
echo "8 is true";
}
echo "<br />";
if (($k & 16) > 0) {
echo "16 is true";
}
echo "<br />";
if (($k & 32) > 0) {
echo "32 is true";
}
When $k is 10 or 11, both 2 and 8 is true.
When $k is 12 or 13, both 4 and 8 is true.
Please help me figure this out

Answer 1

This code is simply checking if certain bit positions are set in some value $k. See how bitwise airthmetic works.
In the code you've sent us, it doesn't have much context. However, this is a typical pattern for checking values in a bitmask.
In general, however, the > 0 is unnecessary (since anything !0 typically == true).
It is important to note that when I say bit positions, I mean of the binary number. That is, $k & 32 does not check bit number 32, it instead checks the 6th most significant bit (32d == 0b100000).
Short example
Consider the last statement. We already determined that decimal 32 looks like 100000 in binary. Now say that we want to check whether or not the 6th bit is set in some number 33. 33 is represented as 0b100001 in binary. Now we take the bitwise and which works much like multiplication that we're used to seeing in base 10 (i.e. decimal).
0b100000 <--- decimal 32
&
0b100001 <--- decimal 33
--------
0b100000 <--- Result is > 0 and, therefore, this value has the 6th bit set
Now, let's retry this example with a different number. How about decimal 15.
0b100000 <--- decimal 32
&
0b001111 <--- decimal 15
--------
0b000000 <--- Result == 0 and, therefore, the 6th bit is not set
Now, this & (not to be confused with logical and which is represented as &&) works for multiple bits as well. Say you want to check for bits 6 and 4, then you would want to use 0b101000 (decimal 40), but depending on what you wanted to test (i.e. exactly those bits or just one of those bits), your conditional may change.
Just a little bit more
From a style standpoint it is important to note that these masks are typically represented in hexadecimal instead of decimal (or binary because that would be cumbersome). This is mostly because it is somewhat unclear what binary representation a decimal number correlates to when looking at it "at-a-glance." Let us again consider decimal 32. In hexadecimal, a binary number can be represented as a hexadecimal number by simply grouping bits by 4's (from right to left). See below
0b100000 <--- decimal 32
0000 === 0x0 (hexadecimal 0)
10 === 0b0010 (implied leading 0's) === 0x2 (hexadecimal 2)
--------
0x20
As you can see, hex 20 is decimal 32 is binary 100000.

Answer 2

In your code there are just some bitwise AND operation which checks if some bits are set or not.
So if we take a look what's going on in the first one:
$k = 48;
if (($k & 2) > 0) {
echo "2 is true";
}
The expression ($k & 2) a bit more readable:
0011 0000 $k
0000 0010 2
------------ &
0000 0000 = 0
So if (0 > 0) is false
And another example:
if (($k & 16) > 0) {
echo "16 is true";
}
The expression ($k & 16) a bit more readable:
0011 0000 $k
0001 0000 16
------------ &
0001 0000 = 16
So if (16 > 0) is true
FYI information:
Bitwise AND operation table:
A | B | Result
--------------------
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1
And as a reference to the manual here: http://php.net/manual/en/language.operators.bitwise.php